Bash – What does “echo (ls)” do in bash

bashecho

When I run echo (ls) in bash it returns:

-bash: syntax error near unexpected token `ls'

I understand that I should escape parenthesis or quote to get plain text output.
But the result still does not make sense to me if parenthesis means running the command sequence in a subshell environment.

My environment: bash 4.3 (installed by homebrew), OS X El Capitan

Best Answer

It is essentially a generic syntax error, not specifically related to the ls token. bash uses a yacc parser, which calls a common yyerror() on any problem, Within the resulting error-handling, it proceeds to try to pinpoint the error. The message is coming from this chunk (see source):

  /* If the line of input we're reading is not null, try to find the       
     objectionable token.  First, try to figure out what token the
     parser's complaining about by looking at current_token. */
  if (current_token != 0 && EOF_Reached == 0 && (msg = error_token_from_token (current_token)))
    {
      if (ansic_shouldquote (msg))
    {
      p = ansic_quote (msg, 0, NULL);
      free (msg);
      msg = p;
    }
      parser_error (line_number, _("syntax error near unexpected token `%s'"), msg);
      free (msg);

      if (interactive == 0)
    print_offending_line ();

      last_command_exit_value = parse_and_execute_level ? EX_BADSYNTAX : EX_BADUSAGE;
      return;
    }

In other words, it's already confused by the '(', and having looked-ahead for context is reporting the ls.

A ( would be legal at the beginning of a command, but not embedded. Per manual page:

   Compound Commands                                                       
       A compound command is one of the following:

       (list) list  is  executed in a subshell environment (see COMMAND EXECU‐
              TION ENVIRONMENT below).  Variable assignments and builtin  com‐
              mands  that  affect  the  shell's  environment  do not remain in
              effect after the command completes.  The return  status  is  the
              exit status of list.

Related Solutions

Bash – Non-printing escape sequence: when

The "non-printing escape sequence" is needed when using non-printing characters in $PS1 because bash needs to know the position of the cursor so that the screen can be updated correctly when you edit the command line. Bash does that by counting the number of characters in the $PS1 prompt and then that's the column number the cursor is in.

However, if you place non-printing sequences in $PS1, then that count is wrong and the line can be messed up if you edit the command line. Hence the introduction of the \[ and \] markers to indicate that the enclosed bytes should not be counted.

Bash – Differences Between Reserved Word and Command time

In bash, time is a reserved word, so the shell can parse it the own way and apply rules for it.

Here is the code show how bash parse line start with time reserved word:

static int
time_command_acceptable ()
{
#if defined (COMMAND_TIMING)
  int i;

  if (posixly_correct && shell_compatibility_level > 41)
    {
      /* Quick check of the rest of the line to find the next token.  If it
     begins with a `-', Posix says to not return `time' as the token.
     This was interp 267. */
      i = shell_input_line_index;
      while (i < shell_input_line_len && (shell_input_line[i] == ' ' || shell_input_line[i] == '\t'))
        i++;
      if (shell_input_line[i] == '-')
    return 0;
    }

  switch (last_read_token)
    {
    case 0:
    case ';':
    case '\n':
    case AND_AND:
    case OR_OR:
    case '&':
    case WHILE:
    case DO:
    case UNTIL:
    case IF:
    case THEN:
    case ELIF:
    case ELSE:
    case '{':       /* } */
    case '(':       /* )( */
    case ')':       /* only valid in case statement */
    case BANG:      /* ! time pipeline */
    case TIME:      /* time time pipeline */
    case TIMEOPT:   /* time -p time pipeline */
    case TIMEIGN:   /* time -p -- ... */
      return 1;
    default:
      return 0;
    }
#else
  return 0;
#endif /* COMMAND_TIMING */
}

You see, time can be followed by most others bash reserved words.

In case of external command, the normal rule was applied, { was considered input of /usr/bin/time. } alone is invalid token, and bash raise the error.

In:

/usr/bin/time echo hello

external time did not call the shell builtin echo but the external echo command.

A strace verifies that:

$ strace -fe execve /usr/bin/time echo 1
execve("/usr/bin/time", ["/usr/bin/time", "echo", "1"], [/* 64 vars */]) = 0
Process 25161 attached
....
[pid 25161] execve("/usr/bin/echo", ["echo", "1"], [/* 64 vars */]) = -1 ENOENT (No such file or directory)
[pid 25161] execve("/bin/echo", ["echo", "1"], [/* 64 vars */]) = 0
1
[pid 25161] +++ exited with 0 +++
....

Here external time lookup your PATH variable to find the command executable. That also explain in case of using a function, you got No such file or directory becasue there's no command named mytest in your PATH.

Best Answer

Related Solutions

Bash – Non-printing escape sequence: when

Bash – Differences Between Reserved Word and Command time

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